Question

If a body of mass \(m\) has to be taken from the surface of earth to a height \(h = R\), then the amount of energy required is (\(R\): radius of earth)

• A. \(mgR\)
• B. \(\frac{mgR}{3}\)
• C. \(\frac{mgR}{2}\)
• D. \(\frac{mgR}{12}\)
• E. \(\frac{mgR}{9}\)

Hint:

For heights comparable to the Earth's radius, do not use \(mgh\). Instead, use the general formula: \(\Delta U = \frac{mgh}{1 + \frac{h}{R}}\). Substituting \(h = R\) gives \(\Delta U = \frac{mgR}{1 + 1} = \frac{mgR}{2}\).

Answer 1

The Correct Option is C

Concept: The energy required to lift an object is equal to the change in its gravitational potential energy. The potential energy at a distance \(r\) from the center of the Earth is given by \(U = -\frac{GMm}{r}\).
• Potential Energy at surface (\(r = R\)): \(U_i = -\frac{GMm}{R}\)
• Potential Energy at height \(h\) (\(r = R+h\)): \(U_f = -\frac{GMm}{R+h}\)
• Relation to \(g\): \(g = \frac{GM}{R^2} \Rightarrow GM = gR^2\)

Step 1: Set up the equation for Change in Potential Energy.
Energy required (\(\Delta U\)) = \(U_f - U_i\). Given \(h = R\), the final distance is \(r = R + R = 2R\). \[ \Delta U = \left( -\frac{GMm}{2R} \right) - \left( -\frac{GMm}{R} \right) \] \[ \Delta U = \frac{GMm}{R} - \frac{GMm}{2R} = \frac{GMm}{2R} \]

Step 2: Substitute \(GM = gR^2\) to get the final form.
\[ \Delta U = \frac{(gR^2)m}{2R} \] \[ \Delta U = \frac{mgR}{2} \]